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gulaghasi [49]
3 years ago
5

A rock weighs 110 N in air and has a volume of 0.00337 m3 . What is its apparent weight when submerged in water? The acceleratio

n of gravity is 9.8 m/s 2 . Answer in units of N.
Physics
1 answer:
pickupchik [31]3 years ago
4 0

Explanation:

It is given that,

Weight of the rock in air, W = 110 N

Since, W = mg

m=\dfrac{W}{g}

m=\dfrac{110\ N}{9.8\ m/s^2}

m = 11.22 kg

We need to find the apparent weight of the rock when it is submerged in water. Apparent weight is equal to the weight of liquid displaced i.e.

M=d\times V

d is the density of water, d=1000\ kg/m^3

V is the volume of rock, V=0.00337\ m^3

M=1000\ kg/m^3\times 0.00337\ m^3

M = 3.37 kg

The apparent weight in water, W = m - M

W=7.85\ kg\times 9.8\ m/s^2

W = 76.93 N

So, the apparent weight of the rock is 76.93 N. Hence, this is the required solution.

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Answer:

The right response will be "450 volts".

Explanation:

The given values are:

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As we know,

The potential difference between the two shell's difference will be:

⇒  \Delta V=K[(\frac{q1}{R1}+\frac{q2}{R2})-(\frac{q1}{R1} +(\frac{q2}{R2}))]

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On substituting the values, we get

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A 100kg crate slides along a floor with a starting velocity of 21 m/s. If the force due to friction is 8N, then, it will take 262.5 s for the box to come to rest.

We'll begin by calculating the declaration of the box. This can be obtained as follow:

Force (F) = –8 N (opposition)

Mass (m) = 100 Kg

<h3>Deceleration (a) =? </h3>

<h3>F = ma</h3>

–8 = 100 × a

Divide both side by 1000

a = \frac{-8}{100}

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Therefore, the deceleration of the box is –0.08 ms¯²

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<h3>Time (t) =.? </h3>

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Collect like terms

0 – 21 = –0.08t

–21 = –0.08t

Divide both side by –0.08

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<h3>t = 262.5 s</h3>

Therefore, it will take 262.5 s for the box to come to rest.

Learn more: brainly.com/question/14446351

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